Math + You = 1

# Hospital's Rule: Statement and Proof

What is Hospital's theorem, also known as Hospital's rule? This article will make it possible to state and demonstrate this rule!

## Hospital rule statement

Let f and g be two differentiable functions at a point a and such that f(a) = g(a) = 0. Moreover, g'(a) ≠ 0. Then, we have the following property:

\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a } \dfrac{f'(x)}{g'(x)}

### Generalization

We can replace a by

\ pm \ infty

And also we can have:

\begin{array}{l} \displaystyle\lim_{x\to a} f(x) = \pm \infty\\ \displaystyle\lim_{x\to a} g(x) = \pm \infty \end {array}

### Examples

Example 1 : Take the function h defined by

h(x) = \dfrac{x-2}{x^2-6x+8} = \dfrac{f(x)}{g(x)}

We have :

f(2)=g(2)=0

On the other hand,

g'(x) = 2x-6

So

g'(2) \neq 0

We then deduce:

\lim_{x \to 2} h(x) = \lim_{x \to 2} \dfrac{1}{2x-6} = -\dfrac{1}{2}

Example 2 : Take the functions f and g defined by

f(x) = \ln(X)

Et

g(x) = x-1

And let's study the limit of the ratio of these two functions in 1.
We have: f(1) = g(1) = 0
Urban artist

g'(x) = 1

So we also have:

g'(1) \neq 0

We will therefore apply our famous rule to obtain:

\lim_{x \to 1}\dfrac{f(1)}{g(1)} = \lim_{x \to 1}\dfrac{f'(1)}{g'(1)} = \dfrac {\frac{1}{1}}{1} = 1

## Demonstration of L'Hôpital's rule

Now let's move on to the demonstration. This is quite simple in the classical case.
We have the following property:

f(x) = f(x)-f(a)

Similarly,

g(x) =g(x)-g(a)

So we do the quotient:

\begin{array}{ll} \dfrac{f(x)}{g(x)} & = \dfrac{f(x)-f(a)}{g(x)-g(a) }\\ & = \dfrac{f(x)-f(a)}{xa } \dfrac{xa}{g(x)-g(a) }\\ \end{array}

Passing to the limit:

\lim_{x\to a }\dfrac{f(x)}{g(x)} =\dfrac{f'(a)}{g'(a)}


This concludes this demonstration of L'Hôpital's rule.

If you are familiar with the usual equivalents and expansions, you will normally never need this rule. But sometimes it can be easy to use and some are fans of it as they master the limited developments!